IB Chemistry HL Reactivity 1: What drives chemical reactions? questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Question 1

Multiple choiceEasy1 mark

The figure shows the alcohol burned in a calorimetry experiment. A 0.230 g sample is combusted at constant pressure, warming 100.0 g of water from 20.0 °C to 26.0 °C. For the estimate, use the heat gained by the water as the heat released by combustion. The specific heat capacity of water is 4.18 J g⁻¹ K⁻¹. What molar enthalpy change of combustion is obtained, in kJ mol⁻¹?

CH₃CH₂OH
Show answer and explanation

Answer: A. The structure gives the formula C₂H₆O, so its molar mass is 46.07 g mol⁻¹. The water gains Q = mcΔT = 100.0 × 4.18 × 6.0 = 2508 J = 2.508 kJ. The amount of ethanol is 0.230/46.07 = 0.00499 mol, giving ΔH = −2.508/0.00499 = −502 kJ mol⁻¹.

B: uses the molar mass of CO₂, 44.0 g mol⁻¹, instead of ethanol’s. C: doubles the ethanol amount by applying the coefficient 2 for CO₂ to the fuel amount. D: reverses the sign, treating heat gained by the water as a positive enthalpy change for combustion.

Remember: heat released by an exothermic reaction gives a negative enthalpy change.

Question 2

Multiple choiceEasy1 mark

A reaction takes place in a vessel immersed in a water bath. The bath cools as the reaction proceeds. Which description of the reaction is correct?

Show answer and explanation

Answer: C. As the bath cools, it loses energy to the reaction. Energy transferred from the surroundings to the reaction makes the reaction endothermic.

A: identifies the reaction as endothermic but reverses the direction of energy transfer.
B: gives the correct direction of transfer but labels energy absorbed by the reaction as exothermic.
D: treats the bath’s cooling as evidence that the reaction releases energy, reversing the transfer direction and classification.
Remember: an endothermic reaction takes in energy from its surroundings.

Question 3

Paper 2Medium3 marks

Citric acid solution is mixed with solid sodium hydrogencarbonate in an insulated cup. Treat the reacting chemicals as the system and the solution and cup as the surroundings. Figure 1 shows the measured temperature of the solution before and after mixing; the reagents are mixed immediately after the reading at t=0t=0

-40-200204060801001201919.52020.52121.52222.523Time / sTemperature / °CReagents mixed immediately after this reading
(a)

Determine the temperature change of the surroundings from immediately before mixing to the lowest recorded value. Include the sign.

[1]
(b)

Hence classify the reaction and explain the temperature change, referring to energy transfer, particle kinetic energy and conservation of energy.

[2]

Question 4

Paper 2Easy4 marks

Ethanol, C₂H₅OH(l), is used as a fuel. Its complete combustion produces carbon dioxide and liquid water. The standard enthalpies of formation at 298 K are:

SubstanceΔH°f / kJ mol⁻¹
C₂H₅OH(l)−277.7
O₂(g)0
CO₂(g)−393.5
H₂O(l)−285.8
(a)

Write the balanced equation for the complete combustion of ethanol, including state symbols.

[2]
(b)

Using the equation in (a) and the equation for enthalpies of formation in the data booklet, determine the standard enthalpy change for the combustion of 1 mol of ethanol. Give your answer in kJ mol⁻¹.

[2]

Question 5

Multiple choiceMedium1 mark

The following standard enthalpy changes are given for a Born–Haber cycle for MgCl₂(s). The chloride electron-affinity step occurs twice. Which value and interpretation of the final step are consistent with the cycle?

ProcessΔH° / kJ mol⁻¹
Mg(s) → Mg(g)+148
Mg(g) → Mg⁺(g) + e⁻+738
Mg⁺(g) → Mg²⁺(g) + e⁻+1451
Cl₂(g) → 2Cl(g)+243
Cl(g) + e⁻ → Cl⁻(g)−349
Mg(s) + Cl₂(g) → MgCl₂(s)−642
Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s)?
Show answer and explanation

Answer: B. The enthalpy changes around the cycle sum to the standard enthalpy of formation: 148 + 738 + 1451 + 243 + 2(−349) + ΔH_latt = −642, so ΔH_latt = −2524 kJ mol⁻¹. The negative value means energy is released as gaseous ions form MgCl₂(s).

A: counts only one electron-affinity step, giving −2873 kJ mol⁻¹.
C: omits the second ionization energy, giving −1073 kJ mol⁻¹.
D: reverses the sign of the lattice step, making formation from gaseous ions endothermic.

Remember: the enthalpy changes around a complete Hess cycle sum to zero.

Question 6

Paper 2Easy8 marks

A student compares the air supply to two identical burners, each supplied with methane, CH₄, at the same rate. Figure 1 is a schematic diagram; the side arrows indicate the relative air supply to each burner.

ABCH4CH4
(a)

Using Figure 1, identify which burner is more likely to undergo incomplete combustion. Give one reason based on the figure.

[2]
(b)

For burner B, write a balanced equation for incomplete combustion of methane in which carbon monoxide and water are the products. State symbols are not required.

[2]
(c)

Ethanol, C₂H₅OH, burns with a restricted oxygen supply to form solid carbon and water. Write a balanced equation for this reaction. State symbols are not required.

[2]
(d)

State one health risk of carbon monoxide and suggest one change that would reduce the likelihood of carbon monoxide forming or accumulating.

[2]

Question 7

Multiple choiceMedium1 mark

A 23.0 g sample of bioethanol, C₂H₅OH, is burned completely. Assume all its carbon was fixed from atmospheric CO₂ by recent photosynthesis, and ignore emissions from production and transport. The combustion equation is C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. Use M(C₂H₅OH) = 46.0 g mol⁻¹ and M(CO₂) = 44.0 g mol⁻¹. What are the mass of CO₂ released and the net change in atmospheric CO₂ from fixation and combustion together? Treat an increase as positive.

Show answer and explanation

Answer: C. The equation gives 2 mol CO₂ per mole of ethanol. The ethanol amount is 23.0/46.0 = 0.500 mol, so combustion releases 1.00 mol CO₂ × 44.0 g mol⁻¹ = 44.0 g. The same 1.00 mol of atmospheric CO₂ had been fixed to provide the carbon in the ethanol, so the net change is +44.0 − 44.0 = 0.0 g.

A: uses a 1:1 rather than 1:2 ethanol-to-CO₂ mole ratio, giving 22.0 g. B: calculates the released CO₂ but omits the earlier photosynthetic uptake. D: counts the uptake as a negative change but omits the CO₂ released on combustion.

Remember: for the stated carbon balance, recently fixed carbon returns to the atmosphere on combustion.

Question 8

Paper 2Medium7 marks

A pilot plant produces bioethanol from glucose in crop biomass. Figure 1 shows a simplified carbon pathway for one batch. The fermentation equation is:

\ceC6H12O6−>2C2H5OH+2CO2\ce{C6H12O6 -> 2C2H5OH + 2CO2}

Use relative atomic masses from the data booklet where needed.

Atmospheric CO2Crop biomass:glucoseFermentationBioethanolphotosynthesis;carbon fixed duringone growing season90.0 kg glucose38.6 kg ethanol collectedCO2 releasedcombustion releasesCO2
(a)

Write a balanced equation for photosynthesis in plants.

[1]
(b)

Use the fermentation equation and Figure 1 to calculate the theoretical mass of ethanol and the percentage yield for this batch. Give the mass in kg and the percentage yield to three significant figures.

[3]
(c)

Using Figure 1, explain why the ethanol can be considered a renewable fuel.

[2]
(d)

State one reason why Figure 1 alone does not establish that using this bioethanol is carbon neutral.

[1]

Question 9

Multiple choiceMedium1 mark
2HOOH(l)HOOH(l)2HOH(l)HOH(l)+OO(g)hydrogen peroxide (SMILES: OO)water (SMILES: O)oxygen (SMILES: O=O)

For the reaction shown, the standard molar entropies are S°(H₂O₂(l)) = 109.6, S°(H₂O(l)) = 69.9 and S°(O₂(g)) = 205.0 J K⁻¹ mol⁻¹. Which option gives ΔᵣS° per mole of reaction as written and the resulting change in the system’s entropy?

Show answer and explanation

Answer: C. For a reaction, ΔᵣS° = ΣνS°(products) − ΣνS°(reactants). Using the coefficients read from the figure: [2(69.9) + 205.0] − [2(109.6)] = +125.6 J K⁻¹ mol⁻¹. The positive value means the system’s entropy increases.

A: reverses products minus reactants, giving −125.6 J K⁻¹ mol⁻¹ and a decrease.
B: omits the coefficient 2 for water, giving +55.7 J K⁻¹ mol⁻¹.
D: omits the coefficient 2 for hydrogen peroxide, giving +235.2 J K⁻¹ mol⁻¹.

Remember: multiply each standard molar entropy by its stoichiometric coefficient before summing.

Question 10

Paper 2Medium7 marks

Calcium carbonate decomposes on heating according to the reaction

\ceCaCO3(s)−>CaO(s)+CO2(g).\ce{CaCO3(s) -> CaO(s) + CO2(g)}.

The standard entropy values and standard enthalpy change for this reaction are given below. Figure 1 shows how ΔrG∘\Delta_\mathrm r G^\circ varies with temperature.

Substance or quantityValue
S∘(\ceCaCO3(s))S^\circ(\ce{CaCO3(s)})92.9 J K−1 mol−192.9\ \mathrm{J\,K^{-1}\,mol^{-1}}
S∘(\ceCaO(s))S^\circ(\ce{CaO(s)})39.8 J K−1 mol−139.8\ \mathrm{J\,K^{-1}\,mol^{-1}}
S∘(\ceCO2(g))S^\circ(\ce{CO2(g)})213.7 J K−1 mol−1213.7\ \mathrm{J\,K^{-1}\,mol^{-1}}
ΔrH∘\Delta_\mathrm r H^\circ+178.3 kJ mol−1+178.3\ \mathrm{kJ\,mol^{-1}}
8009001000110012001300-40-30-20-100102030405060Temperature, T / KΔr G^° / kJ mol-1
(a)

Calculate ΔrS∘\Delta_\mathrm r S^\circ for the reaction. Include the sign and units.

[2]
(b)

Using Figure 1, determine the approximate temperature at which ΔrG∘\Delta_\mathrm r G^\circ changes sign. State whether the forward reaction is thermodynamically favoured under standard-state conditions at 1000 K1000\ \mathrm K

[2]
(c)

Using your answer to (a), calculate ΔrG∘\Delta_\mathrm r G^\circ at 1200 K1200\ \mathrm K State whether the forward reaction is thermodynamically favoured under standard-state conditions at this temperature.

[3]

Question 11

Paper 2Hard12 marks

Calcium oxide is produced by heating limestone, which contains calcium carbonate. The standard Gibbs energy change, ΔrG∘\Delta_\mathrm r G^\circ for the decomposition is shown in Figure 1. Assume that ΔrH∘\Delta_\mathrm r H^\circ and ΔrS∘\Delta_\mathrm r S^\circ remain constant over the temperature range shown.

7008009001000110012001300-40-30-20-10010203040506070T / KΔr G° / kJ mol-1
(a)

Write a balanced equation, including state symbols, for the thermal decomposition of calcium carbonate.

[1]
(b)

Use Figure 1 to estimate ΔrG∘\Delta_\mathrm r G^\circ at 1000 K1000\ \mathrm K

[1]
(c)

Use Figure 1 to determine ΔrH∘\Delta_\mathrm r H^\circ and ΔrS∘\Delta_\mathrm r S^\circ for the decomposition. Give signs and units, and show how you obtain both values.

[3]
(d)

Using your values from (c), determine the temperature at which ΔrG∘=0\Delta_\mathrm r G^\circ=0 State at which temperatures the forward decomposition is spontaneous under standard-state conditions.

[2]
(e)

At 1000 K1000\ \mathrm K a gas stream keeps the carbon dioxide partial pressure in the kiln at 0.10 bar0.10\ \mathrm{bar} For the reaction as written, take

Q=p(CO2)p∘,p∘=1.00 bar,Q=\frac{p(\mathrm{CO_2})}{p^\circ},\qquad p^\circ=1.00\ \mathrm{bar},

with pure solids omitted from QQ Under these conditions,

ΔrG=ΔrG∘+RTln⁡Q,\Delta_\mathrm r G=\Delta_\mathrm r G^\circ+RT\ln Q,

where R=8.31 J K−1 mol−1R=8.31\ \mathrm{J\,K^{-1}\,mol^{-1}}

Determine the carbon dioxide partial pressure at which ΔrG=0\Delta_\mathrm r G=0 at 1000 K1000\ \mathrm K Hence decide whether the forward decomposition is spontaneous at p(CO2)=0.10 barp(\mathrm{CO_2})=0.10\ \mathrm{bar} and assess the claim that a positive ΔrG∘\Delta_\mathrm r G^\circ at 1000 K1000\ \mathrm K means that the decomposition cannot be spontaneous.

[5]

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